No. 21812 is divisible by 2.
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If its divisible by 5 AND 2 it must be divisible by 10 So you just have to pick the only number between 21 and 39 that's divisible by 10
21 is not divisible by 2 nor 5, but it is divisible by 3. To be divisible by 2 the last digit must be even (one of {0, 2, 4, 6, 8}; the last digit 1 is not one of these, thus 21 is not divisible by 2 To be divisible by 3 sum the digits and if this total is divisible by 3, then so is the original number; the test can be repeated on the sum, so by repeatedly summing the digits of the totals until a single digit remains only if that single digit is 3, 6 or 9 is the original number divisible by 3. 21 → 2 + 1 = 3 which is divisible by 3 so 21 is divisible by 3. To be divisible by 5, the last digit must be 0 or 5. The last digit of 21 is 1 which is not 0 nor 5, thus 21 is not divisible by 5.
42 can be divided by 2 once. The result, 21, is not divisible by 2.
The divisibility rule for 3 is as follows: If the sum of the digits of a number is divisible by 3 then the number is divisible by 3. For example, the number 21 is divisible by 3 since 2+1=3, and 3 is divisible by 3. The number 49 is not divisible by 3 since 4+9=13, and 13 is not divisible by 3.
No, it is divisible by: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126